# Maximum XOR of Two Numbers in an Array

• Difficulty Level : Hard
• Last Updated : 17 Nov, 2021

Given an array Arr of non-negative integers of size N. The task is to find the maximum possible xor between two numbers present in the array.
Example

Input: Arr = {25, 10, 2, 8, 5, 3}
Output: 28
The maximum result is 5 ^ 25 = 28
Input: Arr = {1, 2, 3, 4, 5, 6, 7}
Output:
The maximum result is 1 ^ 6 = 7

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Naive Approach: A Simple Solution is to generate all pairs of the given array and compute XOR of the pairs. Finally, return the maximum XOR value. This solution takes time.
Below is the implementation of the above approach:

## C++

 `// C++ implementation``#include ``using` `namespace` `std;` `// Function to return the``// maximum xor``int` `max_xor(``int` `arr[], ``int` `n)``{` `    ``int` `maxXor = 0;` `    ``// Calculating xor of each pair``    ``for` `(``int` `i = 0; i < n; i++) {``        ``for` `(``int` `j = i + 1; j < n; j++) {``            ``maxXor = max(maxXor,``                         ``arr[i] ^ arr[j]);``        ``}``    ``}``    ``return` `maxXor;``}` `// Driver Code``int` `main()``{` `    ``int` `arr[] = { 25, 10, 2, 8, 5, 3 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``cout << max_xor(arr, n) << endl;``    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{` `    ``// Function to return the``    ``// maximum xor``    ``static` `int` `max_xor(``int` `arr[], ``int` `n)``    ``{``        ``int` `maxXor = ``0``;` `        ``// Calculating xor of each pair``        ``for` `(``int` `i = ``0``; i < n; i++)``        ``{``            ``for` `(``int` `j = i + ``1``; j < n; j++)``            ``{``                ``maxXor = Math.max(maxXor,``                        ``arr[i] ^ arr[j]);``            ``}``        ``}``        ``return` `maxXor;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = {``25``, ``10``, ``2``, ``8``, ``5``, ``3``};``        ``int` `n = arr.length;` `        ``System.out.println(max_xor(arr, n));``    ``}``}` `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 implementation` `# Function to return the``# maximum xor``def` `max_xor(arr, n):` `    ``maxXor ``=` `0``;` `    ``# Calculating xor of each pair``    ``for` `i ``in` `range``(n):``        ``for` `j ``in` `range``(i ``+` `1``, n):``            ``maxXor ``=` `max``(maxXor,\``                         ``arr[i] ^ arr[j]);` `    ``return` `maxXor;` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:` `    ``arr ``=` `[ ``25``, ``10``, ``2``, ``8``, ``5``, ``3` `];``    ``n ``=` `len``(arr);` `    ``print``(max_xor(arr, n));` `# This code is contributed by 29AjayKumar`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{` `// Function to return the``// maximum xor``static` `int` `max_xor(``int` `[]arr, ``int` `n)``{``    ``int` `maxXor = 0;` `    ``// Calculating xor of each pair``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``for` `(``int` `j = i + 1; j < n; j++)``        ``{``            ``maxXor = Math.Max(maxXor,``                              ``arr[i] ^ arr[j]);``        ``}``    ``}``    ``return` `maxXor;``}` `// Driver Code``public` `static` `void` `Main()``{``    ``int` `[]arr = {25, 10, 2, 8, 5, 3};``    ``int` `n = arr.Length;` `    ``Console.WriteLine(max_xor(arr, n));``}``}` `// This code is contributed by AnkitRai01`

## Javascript

 ``

Output:
`28`

Time Complexity: , where N is the size of the array

Auxiliary Space: O(1)
Efficient Approach: The approach is similar to this article where the task is to find maximum AND value pair
So the idea is to change the problem statement from finding maximum xor of two numbers in an array to -> find two numbers in an array, such that xor of which equals to a number X. In this case, X will be the maximum number we want to achieve till i-th bit.
To find the largest value of an XOR operation, the value of xor should have every bit to be a set bit i.e 1. In a 32 bit number, the goal is to get the most 1 set starting left to right.
To evaluate each bit, there is a mask needed for that bit. A mask defines which bit should present in the answer and which bit is not. Here we will use mask to keep the prefix for every number ( means by taking the ans with the mask how many bits are remaining from the number ) in the input till the i-th bit then with the list of possible numbers in our set, after inserting the number we will evaluate if we can update the max for that bit position to be 1.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach` `#include ``using` `namespace` `std;` `// Function to return the``// maximum xor``int` `max_xor(``int` `arr[], ``int` `n)``{``    ``int` `maxx = 0, mask = 0;` `    ``set<``int``> se;` `    ``for` `(``int` `i = 30; i >= 0; i--) {` `        ``// set the i'th bit in mask``        ``// like 100000, 110000, 111000..``        ``mask |= (1 << i);` `        ``for` `(``int` `i = 0; i < n; ++i) {` `            ``// Just keep the prefix till``            ``// i'th bit neglecting all``            ``// the bit's after i'th bit``            ``se.insert(arr[i] & mask);``        ``}` `        ``int` `newMaxx = maxx | (1 << i);` `        ``for` `(``int` `prefix : se) {` `            ``// find two pair in set``            ``// such that a^b = newMaxx``            ``// which is the highest``            ``// possible bit can be obtained``            ``if` `(se.count(newMaxx ^ prefix)) {``                ``maxx = newMaxx;``                ``break``;``            ``}``        ``}` `        ``// clear the set for next``        ``// iteration``        ``se.clear();``    ``}` `    ``return` `maxx;``}` `// Driver Code``int` `main()``{` `    ``int` `arr[] = { 25, 10, 2, 8, 5, 3 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``cout << max_xor(arr, n) << endl;` `    ``return` `0;``}`

## Java

 `// Java implementation of the above approach``import` `java.util.*;``class` `GFG``{` `// Function to return the``// maximum xor``static` `int` `max_xor(``int` `arr[], ``int` `n)``{``    ``int` `maxx = ``0``, mask = ``0``;` `    ``HashSet se = ``new` `HashSet();` `    ``for` `(``int` `i = ``30``; i >= ``0``; i--)``    ``{` `        ``// set the i'th bit in mask``        ``// like 100000, 110000, 111000..``        ``mask |= (``1` `<< i);` `        ``for` `(``int` `j = ``0``; j < n; ++j)``        ``{` `            ``// Just keep the prefix till``            ``// i'th bit neglecting all``            ``// the bit's after i'th bit``            ``se.add(arr[j] & mask);``        ``}` `        ``int` `newMaxx = maxx | (``1` `<< i);` `        ``for` `(``int` `prefix : se)``        ``{` `            ``// find two pair in set``            ``// such that a^b = newMaxx``            ``// which is the highest``            ``// possible bit can be obtained``            ``if` `(se.contains(newMaxx ^ prefix))``            ``{``                ``maxx = newMaxx;``                ``break``;``            ``}``        ``}` `        ``// clear the set for next``        ``// iteration``        ``se.clear();``    ``}``    ``return` `maxx;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { ``25``, ``10``, ``2``, ``8``, ``5``, ``3` `};``    ``int` `n = arr.length;` `    ``System.out.println(max_xor(arr, n));``}``}` `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 implementation of the above approach` `# Function to return the``# maximum xor``def` `max_xor( arr , n):``    ` `    ``maxx ``=` `0``    ``mask ``=` `0``;` `    ``se ``=` `set``()``    ` `    ``for` `i ``in` `range``(``30``, ``-``1``, ``-``1``):``        ` `        ``# set the i'th bit in mask``        ``# like 100000, 110000, 111000..``        ``mask |``=` `(``1` `<< i)``        ``newMaxx ``=` `maxx | (``1` `<< i)``    ` `        ``for` `i ``in` `range``(n):``            ` `            ``# Just keep the prefix till``            ``# i'th bit neglecting all``            ``# the bit's after i'th bit``            ``se.add(arr[i] & mask)` `        ``for` `prefix ``in` `se:``            ` `            ``# find two pair in set``            ``# such that a^b = newMaxx``            ``# which is the highest``            ``# possible bit can be obtained``            ``if` `(newMaxx ^ prefix) ``in` `se:``                ``maxx ``=` `newMaxx``                ``break``                ` `        ``# clear the set for next``        ``# iteration``        ``se.clear()``    ``return` `maxx` `# Driver Code``arr ``=` `[ ``25``, ``10``, ``2``, ``8``, ``5``, ``3` `]``n ``=` `len``(arr)``print``(max_xor(arr, n))` `# This code is contributed by ANKITKUMAR34`

## C#

 `// C# implementation of the above approach``using` `System;``using` `System.Collections.Generic;``    ` `class` `GFG``{` `// Function to return the``// maximum xor``static` `int` `max_xor(``int` `[]arr, ``int` `n)``{``    ``int` `maxx = 0, mask = 0;` `    ``HashSet<``int``> se = ``new` `HashSet<``int``>();` `    ``for` `(``int` `i = 30; i >= 0; i--)``    ``{` `        ``// set the i'th bit in mask``        ``// like 100000, 110000, 111000..``        ``mask |= (1 << i);` `        ``for` `(``int` `j = 0; j < n; ++j)``        ``{` `            ``// Just keep the prefix till``            ``// i'th bit neglecting all``            ``// the bit's after i'th bit``            ``se.Add(arr[j] & mask);``        ``}` `        ``int` `newMaxx = maxx | (1 << i);` `        ``foreach` `(``int` `prefix ``in` `se)``        ``{` `            ``// find two pair in set``            ``// such that a^b = newMaxx``            ``// which is the highest``            ``// possible bit can be obtained``            ``if` `(se.Contains(newMaxx ^ prefix))``            ``{``                ``maxx = newMaxx;``                ``break``;``            ``}``        ``}` `        ``// clear the set for next``        ``// iteration``        ``se.Clear();``    ``}``    ``return` `maxx;``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]arr = { 25, 10, 2, 8, 5, 3 };``    ``int` `n = arr.Length;` `    ``Console.WriteLine(max_xor(arr, n));``}``}` `// This code is contributed by Princi Singh`
Output:
`28`

Time Complexity: , where N is the size of the array and M is the maximum number present in the array
Auxiliary Space: O(logM)

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